Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $k = \dfrac{-9p - 54}{2p + 6} \times \dfrac{p^2 + 2p - 3}{-7p^2 + 7p} $
Explanation: First factor the quadratic. $k = \dfrac{-9p - 54}{2p + 6} \times \dfrac{(p + 3)(p - 1)}{-7p^2 + 7p} $ Then factor out any other terms. $k = \dfrac{-9(p + 6)}{2(p + 3)} \times \dfrac{(p + 3)(p - 1)}{-7p(p - 1)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ -9(p + 6) \times (p + 3)(p - 1) } { 2(p + 3) \times -7p(p - 1) } $ $k = \dfrac{ -9(p + 6)(p + 3)(p - 1)}{ -14p(p + 3)(p - 1)} $ Notice that $(p - 1)$ and $(p + 3)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ -9(p + 6)\cancel{(p + 3)}(p - 1)}{ -14p\cancel{(p + 3)}(p - 1)} $ We are dividing by $p + 3$ , so $p + 3 \neq 0$ Therefore, $p \neq -3$ $k = \dfrac{ -9(p + 6)\cancel{(p + 3)}\cancel{(p - 1)}}{ -14p\cancel{(p + 3)}\cancel{(p - 1)}} $ We are dividing by $p - 1$ , so $p - 1 \neq 0$ Therefore, $p \neq 1$ $k = \dfrac{-9(p + 6)}{-14p} $ $k = \dfrac{9(p + 6)}{14p} ; \space p \neq -3 ; \space p \neq 1 $